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For $z\in\mathbb{C}$, let $f(z)=u(z)+iv(z)$ be an entire function such that $f(0)=0$. Is that true that if $$\displaystyle\lim_{|z|\to+\infty}v(z)=0,$$ then $f\equiv 0$?

My attempt was to use the fact that $f$ is entire and therefore is analytic across the complex plane. Therefore, by Cauchy's Integral Formula, taking $\lambda$ a closed and simple path in $\mathbb{C}$, as $f(0)=0$, we have: $$0=f(0)=\frac{1}{2\pi i}\int_\lambda\frac{f(z)}{z-0}dz\Rightarrow\int_\lambda\frac{f(z)}{z}dz=0.$$ But how can I conclude something about $f(z)$ using that $\displaystyle\lim_{|z|\to+\infty}v(z)=0$?

Manatee
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    How about using Liouville's theorem(https://en.wikipedia.org/wiki/Liouville%27s_theorem_(complex_analysis) )? – Kaira Jan 20 '22 at 21:32
  • @Kaira That would be a great ideia, but how can I apply this theorem to my problem? :) Can you explain what you've thought? – Manatee Jan 20 '22 at 21:34
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    “Integer function”? Perhaps you mean “entire function”? – Martin R Jan 20 '22 at 21:35
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    @Manatee How about combining it with this question:https://math.stackexchange.com/questions/2527579/an-entire-function-whose-imaginary-part-is-bounded-is-constant – Kaira Jan 20 '22 at 21:48
  • @Kaira unfortunately I still don't see how these other statements can help me with my question :( – Manatee Jan 21 '22 at 00:11
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    @Manatee The only part left is proving that $v$ is bounded. If $v$ is bounded then by the link I gave you, we can prove the statement. I think you can use the limit equation to prove that it is bounded. – Kaira Jan 21 '22 at 00:22
  • @Kaira now I understand your thought. I'll think about how to show that the imaginary part is bounded. Do you have some idea? – Manatee Jan 21 '22 at 13:31
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    @Manatee Start with the epsilon-delta definition of the limit equation. From that, it should be clear that $v$ is bounded in a certain region. For the other region, use the fact about the maximum/minimum of a continuous function. – Kaira Jan 21 '22 at 18:13

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